Re: SF: Infinity proof
From: Proginoskes (proginoskes_at_email.msn.com)
Date: 04/18/05
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Date: 17 Apr 2005 23:04:36 -0700
jstevh@msn.com wrote:
> [...]
> As other posters have also now been talking about how it's
> impossible to randomly choose out of infinity.
That is not what they're (and we're) saying, but rather it's impossible
to randomly choose out of a countable infinity UNIFORMLY; i.e., so that
any element has just as much of a chance as being picked as any other.
Random just means that it's impossible to know what the result is ahead
of time.
(I)
For instance, you can pick a positive integer at random by choosing a
real number R in the interval [0,1] uniformly (which _can_ be done),
and then choosing the integer as follows:
(1) If R is in (1/2,1] or 0, let N = 1.
(2) Otherwise, let N be the positive integer such that R is in
(1/2^N, 1/2^(N-1)].
The value of N is random, because you don't know what N will be ahead
of time, but not all positive integers are equally likely to be chosen.
The number 3 is twice as likely to be chosen as 4, for instance.
With this distribution (that's the formal way of saying how likely each
outcome is, when choosing at random), the probability of choosing a
positive even number is 1/4 + 1/16 + 1/64 + ... = 1/3, even though
there are just as many positive even integers as positive odd integers.
(II)
The normal way of handling probability concerning positive integers is
to choose numbers the following way:
(1) If N > k, then the probability of choosing N is zero;
(2) Otherwise, the probability of choosing N is 1/k.
This gives a probability distribution depending on k, which is uniform
on {1,2,3,...,k}. (Each of the outcomes 1, 2, ..., k is equally
likely.) Then the probability of a certain propety P is calculated and
called p_k. Then the limit of p_k is taken.
Calculating probabilities with rational numbers can be handled
similarly, once you enumerate (list) them in a particular order.
(III)
Getting back to your "50% claim" ...
The good news is: Yes, there is a probability distribution of the
rational numbers where 50% of the time, your method of choosing factors
results in a non-trivial rational factor of M. In fact you can increase
50% arbitrarily close to 100% (but never actually attaining 100%).
The bad news is: To find this probability distribution, you probably
need to know the prime factors of M ahead of time.
> The problem is solved with my work by human choice, [...]
What exactly is "human choice"? If you can't define it, if it can't be
formalized, then your paper doesn't belong in a mathematical journal.
(Well, a mathematical journal with a good reputation, anyway.)
Does "human choice" involve knowing the factors of M ahead of time? If
so, you've just gone in a circle; you need to factor M in order to use
SF to factor M.
--- Christopher Heckman
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